# Permutations by swapping

Permutations by swapping
You are encouraged to solve this task according to the task description, using any language you may know.

Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.

Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.

Show the permutations and signs of three items, in order of generation here.

Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.

Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.

References

## AutoHotkey

`Permutations_By_Swapping(str, list:=""){	ch := SubStr(str, 1, 1)								; get left-most charachter of str	for i, line in StrSplit(list, "`n")					; for each line in list		loop % StrLen(line) + 1							; loop each possible position			Newlist .= RegExReplace(line, mod(i,2) ? "(?=.{" A_Index-1 "}\$)" : "^.{" A_Index-1 "}\K", ch) "`n"	list := Newlist ? Trim(Newlist, "`n") : ch			; recreate list	if !str := SubStr(str, 2)							; remove charachter from left hand side		return list										; done if str is empty	return Permutations_By_Swapping(str, list)			; else recurse}`
Examples:
`for each, line in StrSplit(Permutations_By_Swapping(1234), "`n")	result .= line "`tSign: " (mod(A_Index,2)? 1 : -1) "`n"MsgBox, 262144, , % resultreturn`
Outputs:
```1234	Sign: 1
1243	Sign: -1
1423	Sign: 1
4123	Sign: -1
4132	Sign: 1
1432	Sign: -1
1342	Sign: 1
1324	Sign: -1
3124	Sign: 1
3142	Sign: -1
3412	Sign: 1
4312	Sign: -1
4321	Sign: 1
3421	Sign: -1
3241	Sign: 1
3214	Sign: -1
2314	Sign: 1
2341	Sign: -1
2431	Sign: 1
4231	Sign: -1
4213	Sign: 1
2413	Sign: -1
2143	Sign: 1
2134	Sign: -1```

## BBC BASIC

`      PROCperms(3)      PRINT      PROCperms(4)      END       DEF PROCperms(n%)      LOCAL p%(), i%, k%, s%      DIM p%(n%)      FOR i% = 1 TO n%        p%(i%) = -i%      NEXT      s% = 1      REPEAT        PRINT "Perm: [ ";        FOR i% = 1 TO n%          PRINT ;ABSp%(i%) " ";        NEXT        PRINT "] Sign: ";s%        k% = 0        FOR i% = 2 TO n%          IF p%(i%)<0 IF ABSp%(i%)>ABSp%(i%-1) IF ABSp%(i%)>ABSp%(k%) k% = i%        NEXT        FOR i% = 1 TO n%-1          IF p%(i%)>0 IF ABSp%(i%)>ABSp%(i%+1) IF ABSp%(i%)>ABSp%(k%) k% = i%        NEXT        IF k% THEN          FOR i% = 1 TO n%            IF ABSp%(i%)>ABSp%(k%) p%(i%) *= -1          NEXT          i% = k%+SGNp%(k%)          SWAP p%(k%),p%(i%)          s% = -s%        ENDIF      UNTIL k% = 0      ENDPROC`
Output:
```Perm: [ 1 2 3 ] Sign: 1
Perm: [ 1 3 2 ] Sign: -1
Perm: [ 3 1 2 ] Sign: 1
Perm: [ 3 2 1 ] Sign: -1
Perm: [ 2 3 1 ] Sign: 1
Perm: [ 2 1 3 ] Sign: -1

Perm: [ 1 2 3 4 ] Sign: 1
Perm: [ 1 2 4 3 ] Sign: -1
Perm: [ 1 4 2 3 ] Sign: 1
Perm: [ 4 1 2 3 ] Sign: -1
Perm: [ 4 1 3 2 ] Sign: 1
Perm: [ 1 4 3 2 ] Sign: -1
Perm: [ 1 3 4 2 ] Sign: 1
Perm: [ 1 3 2 4 ] Sign: -1
Perm: [ 3 1 2 4 ] Sign: 1
Perm: [ 3 1 4 2 ] Sign: -1
Perm: [ 3 4 1 2 ] Sign: 1
Perm: [ 4 3 1 2 ] Sign: -1
Perm: [ 4 3 2 1 ] Sign: 1
Perm: [ 3 4 2 1 ] Sign: -1
Perm: [ 3 2 4 1 ] Sign: 1
Perm: [ 3 2 1 4 ] Sign: -1
Perm: [ 2 3 1 4 ] Sign: 1
Perm: [ 2 3 4 1 ] Sign: -1
Perm: [ 2 4 3 1 ] Sign: 1
Perm: [ 4 2 3 1 ] Sign: -1
Perm: [ 4 2 1 3 ] Sign: 1
Perm: [ 2 4 1 3 ] Sign: -1
Perm: [ 2 1 4 3 ] Sign: 1
Perm: [ 2 1 3 4 ] Sign: -1
```

## C

Implementation of Heap's Algorithm, array length has to be passed as a parameter for non character arrays, as sizeof() will not give correct results when malloc is used. Prints usage on incorrect invocation.

` /*Abhishek Ghosh, 25th October 2017*/ #include<stdlib.h>#include<string.h>#include<stdio.h> int flag = 1; void heapPermute(int n, int arr[],int arrLen){	int temp;	int i; 	if(n==1){		printf("\n["); 		for(i=0;i<arrLen;i++)			printf("%d,",arr[i]);		printf("\b] Sign : %d",flag); 		flag*=-1;	}	else{		for(i=0;i<n-1;i++){			heapPermute(n-1,arr,arrLen); 			if(n%2==0){				temp = arr[i];				arr[i] = arr[n-1];				arr[n-1] = temp;			}			else{				temp = arr[0];				arr[0] = arr[n-1];				arr[n-1] = temp;			}		}		heapPermute(n-1,arr,arrLen);	}} int main(int argC,char* argV[0]){	int *arr, i=0, count = 1;	char* token; 	if(argC==1)		printf("Usage : %s <comma separated list of integers>",argV[0]);	else{		while(argV[1][i]!=00){			if(argV[1][i++]==',')				count++;		} 		arr = (int*)malloc(count*sizeof(int)); 		i = 0; 		token = strtok(argV[1],","); 		while(token!=NULL){			arr[i++] = atoi(token);			token = strtok(NULL,",");		} 		heapPermute(i,arr,count);	} 	return 0;} `

Output:

```C:\rosettaCode>heapPermute.exe 1,2,3

[1,2,3] Sign : 1
[2,1,3] Sign : -1
[3,1,2] Sign : 1
[1,3,2] Sign : -1
[2,3,1] Sign : 1
[3,2,1] Sign : -1
```

## C++

Direct implementation of Johnson-Trotter algorithm from the reference link.

` #include <iostream>#include <vector> using namespace std; vector<int> UpTo(int n, int offset = 0){	vector<int> retval(n);	for (int ii = 0; ii < n; ++ii)		retval[ii] = ii + offset;	return retval;} struct JohnsonTrotterState_{	vector<int> values_;	vector<int> positions_;	// size is n+1, first element is not used	vector<bool> directions_;	int sign_; 	JohnsonTrotterState_(int n) : values_(UpTo(n, 1)), positions_(UpTo(n + 1, -1)), directions_(n + 1, false), sign_(1) {} 	int LargestMobile() const	// returns 0 if no mobile integer exists	{		for (int r = values_.size(); r > 0; --r)		{			const int loc = positions_[r] + (directions_[r] ? 1 : -1);			if (loc >= 0 && loc < values_.size() && values_[loc] < r)				return r;		}		return 0;	} 	bool IsComplete() const { return LargestMobile() == 0; } 	void operator++()	// implement Johnson-Trotter algorithm	{		const int r = LargestMobile();		const int rLoc = positions_[r];		const int lLoc = rLoc + (directions_[r] ? 1 : -1);		const int l = values_[lLoc];		// do the swap		swap(values_[lLoc], values_[rLoc]);		swap(positions_[l], positions_[r]);		sign_ = -sign_;		// change directions		for (auto pd = directions_.begin() + r + 1; pd != directions_.end(); ++pd)			*pd = !*pd;	}}; int main(void){	JohnsonTrotterState_ state(4);	do	{		for (auto v : state.values_)			cout << v << " ";		cout << "\n";		++state;	} while (!state.IsComplete());} `
Output:
```(1 2 3 4 ); sign = 1
(1 2 4 3 ); sign = -1
(1 4 2 3 ); sign = 1
(4 1 2 3 ); sign = -1
(4 1 3 2 ); sign = 1
(1 4 3 2 ); sign = -1
(1 3 4 2 ); sign = 1
(1 3 2 4 ); sign = -1
(3 1 2 4 ); sign = 1
(3 1 4 2 ); sign = -1
(3 4 1 2 ); sign = 1
(4 3 1 2 ); sign = -1
(4 3 2 1 ); sign = 1
(3 4 2 1 ); sign = -1
(3 2 4 1 ); sign = 1
(3 2 1 4 ); sign = -1
(2 3 1 4 ); sign = 1
(2 3 4 1 ); sign = -1
(2 4 3 1 ); sign = 1
(4 2 3 1 ); sign = -1
(4 2 1 3 ); sign = 1
(2 4 1 3 ); sign = -1
(2 1 4 3 ); sign = 1```

## Clojure

### Recursive version

` (defn permutation-swaps  "List of swap indexes to generate all permutations of n elements"  [n]  (if (= n 2) `((0 1))    (let [old-swaps (permutation-swaps (dec n))          swaps-> (partition 2 1 (range n))          swaps<- (reverse swaps->)]      (mapcat (fn [old-swap side]                (case side                  :first swaps<-                  :right (conj swaps<- old-swap)                  :left (conj swaps-> (map inc old-swap))))              (conj old-swaps nil)              (cons :first (cycle '(:left :right)))))))  (defn swap [v [i j]]  (-> v      (assoc i (nth v j))      (assoc j (nth v i))))  (defn permutations [n]  (let [permutations (reduce                       (fn [all-perms new-swap]                         (conj all-perms (swap (last all-perms)                                               new-swap)))                       (vector (vec (range n)))                       (permutation-swaps n))        output (map vector                    permutations                    (cycle '(1 -1)))]    output))  (doseq [n [2 3 4]]  (dorun (map println (permutations n)))) `
Output:
```[[0 1] 1]
[[1 0] -1]
[[0 1 2] 1]
[[0 2 1] -1]
[[2 0 1] 1]
[[2 1 0] -1]
[[1 2 0] 1]
[[1 0 2] -1]
[[0 1 2 3] 1]
[[0 1 3 2] -1]
[[0 3 1 2] 1]
[[3 0 1 2] -1]
[[3 0 2 1] 1]
[[0 3 2 1] -1]
[[0 2 3 1] 1]
[[0 2 1 3] -1]
[[2 0 1 3] 1]
[[2 0 3 1] -1]
[[2 3 0 1] 1]
[[3 2 0 1] -1]
[[3 2 1 0] 1]
[[2 3 1 0] -1]
[[2 1 3 0] 1]
[[2 1 0 3] -1]
[[1 2 0 3] 1]
[[1 2 3 0] -1]
[[1 3 2 0] 1]
[[3 1 2 0] -1]
[[3 1 0 2] 1]
[[1 3 0 2] -1]
[[1 0 3 2] 1]
[[1 0 2 3] -1]
```

### Modeled After Python version

Translation of: Python
` (ns test-p.core) (defn numbers-only [x]  " Just shows the numbers only for the pairs (i.e. drops the direction --used for display purposes when printing the result"  (mapv first x)) (defn next-permutation  " Generates next permutation from the current (p) using the Johnson-Trotter technique    The code below translates the Python version which has the following steps:     p of form [...[n dir]...] such as [[0 1] [1 1] [2 -1]], where n is a number and dir = direction (=1=right, -1=left, 0=don't move)     Step: 1 finds the pair [n dir] with the largest value of n (where dir is not equal to 0 (done if none)     Step: 2: swap the max pair found with its neighbor in the direction of the pair (i.e. +1 means swap to right, -1 means swap left     Step 3: if swapping places the pair a the beginning or end of the list, set the direction = 0 (i.e. becomes non-mobile)     Step 4: Set the directions of all pairs whose numbers are greater to the right of where the pair was moved to -1 and to the left to +1 "  [p]  (if (every? zero? (map second p))    nil                                                                 ; no mobile elements (all directions are zero)    (let [n (count p)          ; Step 1          fn-find-max (fn [m]                        (first (apply max-key                           ; find the max mobile elment                                   (fn [[i x]]                                     (if (zero? (second x))                                       -1                                       (first x)))                                              (map-indexed vector p))))          i1 (fn-find-max p)                                            ; index of max          [n1 d1] (p i1)                                                ; value and direction of max          i2 (+ d1 i1)          fn-swap (fn [m] (assoc m i2 (m i1) i1 (m i2)))                ; function to swap with neighbor in our step direction          fn-update-max (fn [m] (if (or (contains? #{0 (dec n)} i2)     ; update direction of max (where max went)                                        (> ((m (+ i2 d1)) 0) n1))                                  (assoc-in m [i2 1] 0)                                  m))          fn-update-others (fn [[i3 [n3 d3]]]                            ; Updates directions of pairs to the left and right of max                             (cond                                       ; direction reset to -1 if to right, +1 if to left                               (<= n3 n1) [n3 d3]                               (< i3 i2) [n3 1]                               :else      [n3 -1]))]      ; apply steps 2, 3, 4(using functions that where created for these steps)      (mapv fn-update-others (map-indexed vector (fn-update-max (fn-swap p))))))) (defn spermutations  " Lazy sequence of permutations of n digits"  ; Each element is two element vector (number direction)  ; Startup case - generates sequence 0...(n-1) with move direction (1 = move right, -1 = move left, 0 = don't move)  ([n] (spermutations 1                      (into [] (for [i (range n)] (if (zero? i)                                                    [i 0]               ; 0th element is not mobile yet                                                    [i -1])))))         ; all others move left  ([sign p]   (when-let [s (seq p)]             (cons [(numbers-only p) sign]                   (spermutations (- sign) (next-permutation p))))))   ; recursively tag onto sequence  ;; Print results for 2, 3, and 4 items(doseq [n (range 2 5)]  (do    (println)    (println (format "Permutations and sign of %d items " n))  (doseq [q (spermutations n)] (println (format "Perm: %s Sign: %2d" (first q) (second q)))))) `
Output:
```Permutations and sign of 2 items
Perm: [0 1] Sign:  1
Perm: [1 0] Sign: -1

Permutations and sign of 3 items
Perm: [0 1 2] Sign:  1
Perm: [0 2 1] Sign: -1
Perm: [2 0 1] Sign:  1
Perm: [2 1 0] Sign: -1
Perm: [1 2 0] Sign:  1
Perm: [1 0 2] Sign: -1

Permutations and sign of 4 items
Perm: [0 1 2 3] Sign:  1
Perm: [0 1 3 2] Sign: -1
Perm: [0 3 1 2] Sign:  1
Perm: [3 0 1 2] Sign: -1
Perm: [3 0 2 1] Sign:  1
Perm: [0 3 2 1] Sign: -1
Perm: [0 2 3 1] Sign:  1
Perm: [0 2 1 3] Sign: -1
Perm: [2 0 1 3] Sign:  1
Perm: [2 0 3 1] Sign: -1
Perm: [2 3 0 1] Sign:  1
Perm: [3 2 0 1] Sign: -1
Perm: [3 2 1 0] Sign:  1
Perm: [2 3 1 0] Sign: -1
Perm: [2 1 3 0] Sign:  1
Perm: [2 1 0 3] Sign: -1
Perm: [1 2 0 3] Sign:  1
Perm: [1 2 3 0] Sign: -1
Perm: [1 3 2 0] Sign:  1
Perm: [3 1 2 0] Sign: -1
Perm: [3 1 0 2] Sign:  1
Perm: [1 3 0 2] Sign: -1
Perm: [1 0 3 2] Sign:  1
Perm: [1 0 2 3] Sign: -1
```

## Common Lisp

`(defstruct (directed-number (:conc-name dn-))  (number nil :type integer)  (direction nil :type (member :left :right))) (defmethod print-object ((dn directed-number) stream)  (ecase (dn-direction dn)    (:left  (format stream "<~D" (dn-number dn)))    (:right (format stream "~D>" (dn-number dn))))) (defun dn> (dn1 dn2)  (declare (directed-number dn1 dn2))  (> (dn-number dn1) (dn-number dn2))) (defun dn-reverse-direction (dn)  (declare (directed-number dn))  (setf (dn-direction dn) (ecase (dn-direction dn)                            (:left  :right)                            (:right :left)))) (defun make-directed-numbers-upto (upto)  (let ((numbers (make-array upto :element-type 'integer)))    (dotimes (n upto numbers)      (setf (aref numbers n) (make-directed-number :number (1+ n) :direction :left))))) (defun max-mobile-pos (numbers)  (declare ((vector directed-number) numbers))  (loop with pos-limit = (1- (length numbers))        with max-value and max-pos        for num across numbers        for pos from 0        do (ecase (dn-direction num)             (:left  (when (and (plusp pos) (dn> num (aref numbers (1- pos)))                                (or (null max-value) (dn> num max-value)))                       (setf max-value num                             max-pos   pos)))             (:right (when (and (< pos pos-limit) (dn> num (aref numbers (1+ pos)))                                (or (null max-value) (dn> num max-value)))                       (setf max-value num                             max-pos   pos))))        finally (return max-pos))) (defun permutations (upto)  (loop with numbers = (make-directed-numbers-upto upto)        for max-mobile-pos = (max-mobile-pos numbers)        for sign = 1 then (- sign)        do (format t "~A sign: ~:[~;+~]~D~%" numbers (plusp sign) sign)        while max-mobile-pos        do (let ((max-mobile-number (aref numbers max-mobile-pos)))             (ecase (dn-direction max-mobile-number)               (:left  (rotatef (aref numbers (1- max-mobile-pos))                                (aref numbers max-mobile-pos)))               (:right (rotatef (aref numbers max-mobile-pos)                                (aref numbers (1+ max-mobile-pos)))))             (loop for n across numbers                   when (dn> n max-mobile-number)                     do (dn-reverse-direction n))))) (permutations 3)(permutations 4)`
Output:
```#(<1 <2 <3) sign: +1
#(<1 <3 <2) sign: -1
#(<3 <1 <2) sign: +1
#(3> <2 <1) sign: -1
#(<2 3> <1) sign: +1
#(<2 <1 3>) sign: -1
#(<1 <2 <3 <4) sign: +1
#(<1 <2 <4 <3) sign: -1
#(<1 <4 <2 <3) sign: +1
#(<4 <1 <2 <3) sign: -1
#(4> <1 <3 <2) sign: +1
#(<1 4> <3 <2) sign: -1
#(<1 <3 4> <2) sign: +1
#(<1 <3 <2 4>) sign: -1
#(<3 <1 <2 <4) sign: +1
#(<3 <1 <4 <2) sign: -1
#(<3 <4 <1 <2) sign: +1
#(<4 <3 <1 <2) sign: -1
#(4> 3> <2 <1) sign: +1
#(3> 4> <2 <1) sign: -1
#(3> <2 4> <1) sign: +1
#(3> <2 <1 4>) sign: -1
#(<2 3> <1 <4) sign: +1
#(<2 3> <4 <1) sign: -1
#(<2 <4 3> <1) sign: +1
#(<4 <2 3> <1) sign: -1
#(4> <2 <1 3>) sign: +1
#(<2 4> <1 3>) sign: -1
#(<2 <1 4> 3>) sign: +1
#(<2 <1 3> 4>) sign: -1```

## D

### Iterative Version

This isn't a Range yet.

Translation of: Python
`import std.algorithm, std.array, std.typecons, std.range; struct Spermutations(bool doCopy=true) {    private immutable uint n;    alias TResult = Tuple!(int[], int);     int opApply(in int delegate(in ref TResult) nothrow dg) nothrow {        int result;         int sign = 1;        alias Int2 = Tuple!(int, int);        auto p = n.iota.map!(i => Int2(i, i ? -1 : 0)).array;        TResult aux;         aux[0] = p.map!(pi => pi[0]).array;        aux[1] = sign;        result = dg(aux);        if (result)            goto END;         while (p.any!q{ a[1] }) {            // Failed to use std.algorithm here, too much complex.            auto largest = Int2(-100, -100);            int i1 = -1;            foreach (immutable i, immutable pi; p)                if (pi[1])                    if (pi[0] > largest[0]) {                        i1 = i;                        largest = pi;                    }            immutable n1 = largest[0],                      d1 = largest[1];             sign *= -1;            int i2;            if (d1 == -1) {                i2 = i1 - 1;                p[i1].swap(p[i2]);                if (i2 == 0 || p[i2 - 1][0] > n1)                    p[i2][1] = 0;            } else if (d1 == 1) {                i2 = i1 + 1;                p[i1].swap(p[i2]);                if (i2 == n - 1 || p[i2 + 1][0] > n1)                    p[i2][1] = 0;            }             if (doCopy) {                aux[0] = p.map!(pi => pi[0]).array;            } else {                foreach (immutable i, immutable pi; p)                    aux[0][i] = pi[0];            }            aux[1] = sign;            result = dg(aux);            if (result)                goto END;             foreach (immutable i3, ref pi; p) {                immutable n3 = pi[0],                          d3 = pi[1];                if (n3 > n1)                    pi[1] = (i3 < i2) ? 1 : -1;            }        }         END: return result;    }} Spermutations!doCopy spermutations(bool doCopy=true)(in uint n) {    return typeof(return)(n);} version (permutations_by_swapping1) {    void main() {        import std.stdio;        foreach (immutable n; [3, 4]) {            writefln("\nPermutations and sign of %d items", n);            foreach (const tp; n.spermutations)                writefln("Perm: %s  Sign: %2d", tp[]);        }    }}`

Compile with version=permutations_by_swapping1 to see the demo output.

Output:
```Permutations and sign of 3 items
Perm: [0, 1, 2]  Sign:  1
Perm: [0, 2, 1]  Sign: -1
Perm: [2, 0, 1]  Sign:  1
Perm: [2, 1, 0]  Sign: -1
Perm: [1, 2, 0]  Sign:  1
Perm: [1, 0, 2]  Sign: -1

Permutations and sign of 4 items
Perm: [0, 1, 2, 3]  Sign:  1
Perm: [0, 1, 3, 2]  Sign: -1
Perm: [0, 3, 1, 2]  Sign:  1
Perm: [3, 0, 1, 2]  Sign: -1
Perm: [3, 0, 2, 1]  Sign:  1
Perm: [0, 3, 2, 1]  Sign: -1
Perm: [0, 2, 3, 1]  Sign:  1
Perm: [0, 2, 1, 3]  Sign: -1
Perm: [2, 0, 1, 3]  Sign:  1
Perm: [2, 0, 3, 1]  Sign: -1
Perm: [2, 3, 0, 1]  Sign:  1
Perm: [3, 2, 0, 1]  Sign: -1
Perm: [3, 2, 1, 0]  Sign:  1
Perm: [2, 3, 1, 0]  Sign: -1
Perm: [2, 1, 3, 0]  Sign:  1
Perm: [2, 1, 0, 3]  Sign: -1
Perm: [1, 2, 0, 3]  Sign:  1
Perm: [1, 2, 3, 0]  Sign: -1
Perm: [1, 3, 2, 0]  Sign:  1
Perm: [3, 1, 2, 0]  Sign: -1
Perm: [3, 1, 0, 2]  Sign:  1
Perm: [1, 3, 0, 2]  Sign: -1
Perm: [1, 0, 3, 2]  Sign:  1
Perm: [1, 0, 2, 3]  Sign: -1```

### Recursive Version

Translation of: Python
`import std.algorithm, std.array, std.typecons, std.range; auto sPermutations(in uint n) pure nothrow @safe {    static immutable(int[])[] inner(in int items) pure nothrow @safe {        if (items <= 0)            return [[]];        typeof(return) r;        foreach (immutable i, immutable item; inner(items - 1)) {            //r.put((i % 2 ? iota(item.length.signed, -1, -1) :            //               iota(item.length + 1))            //      .map!(i => item[0 .. i] ~ (items - 1) ~ item[i .. \$]));            immutable f = (in size_t i) pure nothrow @safe =>                item[0 .. i] ~ (items - 1) ~ item[i .. \$];            r ~= (i % 2) ?                 //iota(item.length.signed, -1, -1).map!f.array :                 iota(item.length + 1).retro.map!f.array :                 iota(item.length + 1).map!f.array;        }        return r;    }     return inner(n).zip([1, -1].cycle);} void main() {    import std.stdio;    foreach (immutable n; [2, 3, 4]) {        writefln("Permutations and sign of %d items:", n);        foreach (immutable tp; n.sPermutations)            writefln("  %s Sign: %2d", tp[]);        writeln;    }}`
Output:
```Permutations and sign of 2 items:
[1, 0] Sign:  1
[0, 1] Sign: -1

Permutations and sign of 3 items:
[2, 1, 0] Sign:  1
[1, 2, 0] Sign: -1
[1, 0, 2] Sign:  1
[0, 1, 2] Sign: -1
[0, 2, 1] Sign:  1
[2, 0, 1] Sign: -1

Permutations and sign of 4 items:
[3, 2, 1, 0] Sign:  1
[2, 3, 1, 0] Sign: -1
[2, 1, 3, 0] Sign:  1
[2, 1, 0, 3] Sign: -1
[1, 2, 0, 3] Sign:  1
[1, 2, 3, 0] Sign: -1
[1, 3, 2, 0] Sign:  1
[3, 1, 2, 0] Sign: -1
[3, 1, 0, 2] Sign:  1
[1, 3, 0, 2] Sign: -1
[1, 0, 3, 2] Sign:  1
[1, 0, 2, 3] Sign: -1
[0, 1, 2, 3] Sign:  1
[0, 1, 3, 2] Sign: -1
[0, 3, 1, 2] Sign:  1
[3, 0, 1, 2] Sign: -1
[3, 0, 2, 1] Sign:  1
[0, 3, 2, 1] Sign: -1
[0, 2, 3, 1] Sign:  1
[0, 2, 1, 3] Sign: -1
[2, 0, 1, 3] Sign:  1
[2, 0, 3, 1] Sign: -1
[2, 3, 0, 1] Sign:  1
[3, 2, 0, 1] Sign: -1
```

## EchoLisp

The function (in-permutations n) returns a stream which delivers permutations according to the Steinhaus–Johnson–Trotter algorithm.

` (lib 'list) (for/fold (sign 1) ((σ (in-permutations 4)) (count 100))     (printf "perm: %a count:%4d sign:%4d" σ count sign) (* sign -1)) perm: (0 1 2 3) count:   0 sign:   1perm: (0 1 3 2) count:   1 sign:  -1perm: (0 3 1 2) count:   2 sign:   1perm: (3 0 1 2) count:   3 sign:  -1perm: (3 0 2 1) count:   4 sign:   1perm: (0 3 2 1) count:   5 sign:  -1perm: (0 2 3 1) count:   6 sign:   1perm: (0 2 1 3) count:   7 sign:  -1perm: (2 0 1 3) count:   8 sign:   1perm: (2 0 3 1) count:   9 sign:  -1perm: (2 3 0 1) count:  10 sign:   1perm: (3 2 0 1) count:  11 sign:  -1perm: (3 2 1 0) count:  12 sign:   1perm: (2 3 1 0) count:  13 sign:  -1perm: (2 1 3 0) count:  14 sign:   1perm: (2 1 0 3) count:  15 sign:  -1perm: (1 2 0 3) count:  16 sign:   1perm: (1 2 3 0) count:  17 sign:  -1perm: (1 3 2 0) count:  18 sign:   1perm: (3 1 2 0) count:  19 sign:  -1perm: (3 1 0 2) count:  20 sign:   1perm: (1 3 0 2) count:  21 sign:  -1perm: (1 0 3 2) count:  22 sign:   1perm: (1 0 2 3) count:  23 sign:  -1 `

## Elixir

Translation of: Ruby
`defmodule Permutation do  def by_swap(n) do    p = Enum.to_list(0..-n) |> List.to_tuple    by_swap(n, p, 1)  end   defp by_swap(n, p, s) do    IO.puts "Perm: #{inspect for i <- 1..n, do: abs(elem(p,i))}  Sign: #{s}"    k = 0 |> step_up(n, p) |> step_down(n, p)    if k > 0 do      pk = elem(p,k)      i = if pk>0, do: k+1, else: k-1      p = Enum.reduce(1..n, p, fn i,acc ->        if abs(elem(p,i)) > abs(pk), do: put_elem(acc, i, -elem(acc,i)), else: acc      end)      pi = elem(p,i)      p = put_elem(p,i,pk) |> put_elem(k,pi)            # swap      by_swap(n, p, -s)    end  end   defp step_up(k, n, p) do    Enum.reduce(2..n, k, fn i,acc ->      if elem(p,i)<0 and abs(elem(p,i))>abs(elem(p,i-1)) and abs(elem(p,i))>abs(elem(p,acc)),        do: i, else: acc     end)  end   defp step_down(k, n, p) do    Enum.reduce(1..n-1, k, fn i,acc ->      if elem(p,i)>0 and abs(elem(p,i))>abs(elem(p,i+1)) and abs(elem(p,i))>abs(elem(p,acc)),        do: i, else: acc     end)  endend Enum.each(3..4, fn n ->  Permutation.by_swap(n)  IO.puts ""end)`
Output:
```Perm: [1, 2, 3]  Sign: 1
Perm: [1, 3, 2]  Sign: -1
Perm: [3, 1, 2]  Sign: 1
Perm: [3, 2, 1]  Sign: -1
Perm: [2, 3, 1]  Sign: 1
Perm: [2, 1, 3]  Sign: -1

Perm: [1, 2, 3, 4]  Sign: 1
Perm: [1, 2, 4, 3]  Sign: -1
Perm: [1, 4, 2, 3]  Sign: 1
Perm: [4, 1, 2, 3]  Sign: -1
Perm: [4, 1, 3, 2]  Sign: 1
Perm: [1, 4, 3, 2]  Sign: -1
Perm: [1, 3, 4, 2]  Sign: 1
Perm: [1, 3, 2, 4]  Sign: -1
Perm: [3, 1, 2, 4]  Sign: 1
Perm: [3, 1, 4, 2]  Sign: -1
Perm: [3, 4, 1, 2]  Sign: 1
Perm: [4, 3, 1, 2]  Sign: -1
Perm: [4, 3, 2, 1]  Sign: 1
Perm: [3, 4, 2, 1]  Sign: -1
Perm: [3, 2, 4, 1]  Sign: 1
Perm: [3, 2, 1, 4]  Sign: -1
Perm: [2, 3, 1, 4]  Sign: 1
Perm: [2, 3, 4, 1]  Sign: -1
Perm: [2, 4, 3, 1]  Sign: 1
Perm: [4, 2, 3, 1]  Sign: -1
Perm: [4, 2, 1, 3]  Sign: 1
Perm: [2, 4, 1, 3]  Sign: -1
Perm: [2, 1, 4, 3]  Sign: 1
Perm: [2, 1, 3, 4]  Sign: -1
```

## F#

See [2] for an example using this module

` (*Implement Johnson-Trotter algorithm  Nigel Galloway January 24th 2017*)module Ringlet PlainChanges (N:'n[]) = seq{  let gn  = [|for n in N -> 1|]  let ni  = [|for n in N -> 0|]  let gel = Array.length(N)-1  yield Some N  let rec _Ni g e l = seq{    match (l,g) with    |_ when l<0   -> gn.[g] <- -gn.[g]; yield! _Ni (g-1) e (ni.[g-1] + gn.[g-1])    |(1,0)        -> yield None    |_ when l=g+1 -> gn.[g] <- -gn.[g]; yield! _Ni (g-1) (e+1) (ni.[g-1] + gn.[g-1])    |_ -> let n = N.[g-ni.[g]+e];          N.[g-ni.[g]+e] <- N.[g-l+e]; N.[g-l+e] <- n; yield Some N          ni.[g] <- l; yield! _Ni gel 0 (ni.[gel] + gn.[gel])}  yield! _Ni gel 0 1}`

A little code for the purpose of this task demonstrating the algorithm

` for n in Ring.PlainChanges [|1;2;3;4|] do printfn "%A" n `
Output:
```Some [|1; 2; 3; 4|]
Some [|1; 2; 4; 3|]
Some [|1; 4; 2; 3|]
Some [|4; 1; 2; 3|]
Some [|4; 1; 3; 2|]
Some [|1; 4; 3; 2|]
Some [|1; 3; 4; 2|]
Some [|1; 3; 2; 4|]
Some [|3; 1; 2; 4|]
Some [|3; 1; 4; 2|]
Some [|3; 4; 1; 2|]
Some [|4; 3; 1; 2|]
Some [|4; 3; 2; 1|]
Some [|3; 4; 2; 1|]
Some [|3; 2; 4; 1|]
Some [|3; 2; 1; 4|]
Some [|2; 3; 1; 4|]
Some [|2; 3; 4; 1|]
Some [|2; 4; 3; 1|]
Some [|4; 2; 3; 1|]
Some [|4; 2; 1; 3|]
Some [|2; 4; 1; 3|]
Some [|2; 1; 4; 3|]
Some [|2; 1; 3; 4|]
<null>
```

## Forth

Works with: gforth version 0.7.9_20170308
Translation of: BBC BASIC
`S" fsl-util.fs" REQUIREDS" fsl/dynmem.seq" REQUIRED cell darray p{ : sgn  DUP 0 > IF    DROP 1  ELSE 0 < IF    -1  ELSE    0  THEN THEN ;: arr-swap {: addr1 addr2 | tmp -- :}  addr1 @ TO tmp  addr2 @ addr1 !  tmp addr2 ! ;: perms {: n xt | my-i k s -- :}  & p{ n 1+ }malloc malloc-fail? ABORT" perms :: out of memory"  0 p{ 0 } !  n 1+ 1 DO    I NEGATE p{ I } !  LOOP  1 TO s  BEGIN    1 n 1+ DO      p{ I } @ ABS    -1 +LOOP    n 1+ s xt EXECUTE    0 TO k    n 1+ 2 DO      p{ I } @ 0 < ( flag )      p{ I } @ ABS  p{ I 1- } @ ABS  > ( flag flag )      p{ I } @ ABS p{ k } @ ABS > ( flag flag flag )      AND AND IF        I TO k      THEN    LOOP    n 1 DO      p{ I } @ 0 > ( flag )      p{ I } @ ABS  p{ I 1+ } @ ABS  > ( flag flag )      p{ I } @ ABS  p{ k } @ ABS  > ( flag flag flag )      AND AND IF        I TO k      THEN    LOOP    k IF      n 1+ 1 DO        p{ I } @ ABS  p{ k } @ ABS  > IF          p{ I } @ NEGATE p{ I } !        THEN      LOOP      p{ k } @ sgn k + TO my-i      p{ k } p{ my-i } arr-swap      s NEGATE TO s    THEN  k 0 = UNTIL ;: .perm ( p0 p1 p2 ... pn n s )  >R  ." Perm: [ "  1 DO    . SPACE  LOOP  R> ." ] Sign: " . CR ; 3 ' .perm perms CR4 ' .perm perms`

## FreeBASIC

Translation of: BBC BASIC
`' version 31-03-2017' compile with: fbc -s console Sub perms(n As ULong)     Dim As Long p(n), i, k, s = 1     For i = 1 To n        p(i) = -i    Next     Do        Print "Perm: [ ";        For i = 1 To n            Print Abs(p(i)); " ";        Next        Print "] Sign: "; s         k = 0        For i = 2 To n            If p(i) < 0 Then                If Abs(p(i)) > Abs(p(i -1)) Then                    If Abs(p(i)) > Abs(p(k)) Then k = i                End If            End If        Next         For i = 1 To n -1            If p(i) > 0 Then                If Abs(p(i)) > Abs(p(i +1)) Then                    If Abs(p(i)) > Abs(p(k)) Then k = i                End If            End If        Next         If k Then            For  i = 1 To n                If Abs(p(i)) > Abs(p(k)) Then p(i) = -p(i)            Next            i = k + Sgn(p(k))            Swap p(k), p(i)            s = -s        End If     Loop Until k = 0 End Sub ' ------=< MAIN >=------ perms(3)printperms(4) ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd `
Output:
```output is edited to show results side by side
Perm: [  1  2  3 ] Sign:  1         Perm: [  1  2  3  4 ] Sign:  1
Perm: [  1  3  2 ] Sign: -1         Perm: [  1  2  4  3 ] Sign: -1
Perm: [  3  1  2 ] Sign:  1         Perm: [  1  4  2  3 ] Sign:  1
Perm: [  3  2  1 ] Sign: -1         Perm: [  4  1  2  3 ] Sign: -1
Perm: [  2  3  1 ] Sign:  1         Perm: [  4  1  3  2 ] Sign:  1
Perm: [  2  1  3 ] Sign: -1         Perm: [  1  4  3  2 ] Sign: -1
Perm: [  1  2  3  4 ] Sign:  1
Perm: [  1  2  4  3 ] Sign: -1
Perm: [  1  4  2  3 ] Sign:  1
Perm: [  4  1  2  3 ] Sign: -1
Perm: [  4  1  3  2 ] Sign:  1
Perm: [  1  4  3  2 ] Sign: -1
Perm: [  1  3  4  2 ] Sign:  1
Perm: [  1  3  2  4 ] Sign: -1
Perm: [  3  1  2  4 ] Sign:  1
Perm: [  3  1  4  2 ] Sign: -1
Perm: [  3  4  1  2 ] Sign:  1
Perm: [  4  3  1  2 ] Sign: -1
Perm: [  4  3  2  1 ] Sign:  1
Perm: [  3  4  2  1 ] Sign: -1
Perm: [  3  2  4  1 ] Sign:  1
Perm: [  3  2  1  4 ] Sign: -1
Perm: [  2  3  1  4 ] Sign:  1
Perm: [  2  3  4  1 ] Sign: -1
Perm: [  2  4  3  1 ] Sign:  1
Perm: [  4  2  3  1 ] Sign: -1
Perm: [  4  2  1  3 ] Sign:  1
Perm: [  2  4  1  3 ] Sign: -1
Perm: [  2  1  4  3 ] Sign:  1
Perm: [  2  1  3  4 ] Sign: -1```

## Go

`package permute // Iter takes a slice p and returns an iterator function.  The iterator// permutes p in place and returns the sign.  After all permutations have// been generated, the iterator returns 0 and p is left in its initial order.func Iter(p []int) func() int {    f := pf(len(p))    return func() int {        return f(p)    }} // Recursive function used by perm, returns a chain of closures that// implement a loopless recursive SJT.func pf(n int) func([]int) int {    sign := 1    switch n {    case 0, 1:        return func([]int) (s int) {            s = sign            sign = 0            return        }    default:        p0 := pf(n - 1)        i := n        var d int        return func(p []int) int {            switch {            case sign == 0:            case i == n:                i--                sign = p0(p[:i])                d = -1            case i == 0:                i++                sign *= p0(p[1:])                d = 1                if sign == 0 {                    p[0], p[1] = p[1], p[0]                }            default:                p[i], p[i-1] = p[i-1], p[i]                sign = -sign                i += d            }            return sign        }    }}`
`package main import (    "fmt"    "permute") func main() {    p := []int{11, 22, 33}    i := permute.Iter(p)    for sign := i(); sign != 0; sign = i() {        fmt.Println(p, sign)    }}`
Output:
```[11 22 33] 1
[11 33 22] -1
[33 11 22] 1
[33 22 11] -1
[22 33 11] 1
[22 11 33] -1
```

`sPermutations :: [a] -> [([a], Int)]sPermutations = flip zip (cycle [-1, 1]) . foldr aux [[]]  where    aux x items = do      (f, item) <- zip (repeat id) items      f (insertEv x item)    insertEv x [] = [[x]]    insertEv x l@(y:ys) = (x : l) : ((y :) <\$> insertEv x ys) main :: IO ()main = do  putStrLn "3 items:"  mapM_ print \$ sPermutations [1 .. 3]  putStrLn "\n4 items:"  mapM_ print \$ sPermutations [1 .. 4]`
Output:
```3 items:
([1,2,3],-1)
([2,1,3],1)
([2,3,1],-1)
([1,3,2],1)
([3,1,2],-1)
([3,2,1],1)

4 items:
([1,2,3,4],-1)
([2,1,3,4],1)
([2,3,1,4],-1)
([2,3,4,1],1)
([1,3,2,4],-1)
([3,1,2,4],1)
([3,2,1,4],-1)
([3,2,4,1],1)
([1,3,4,2],-1)
([3,1,4,2],1)
([3,4,1,2],-1)
([3,4,2,1],1)
([1,2,4,3],-1)
([2,1,4,3],1)
([2,4,1,3],-1)
([2,4,3,1],1)
([1,4,2,3],-1)
([4,1,2,3],1)
([4,2,1,3],-1)
([4,2,3,1],1)
([1,4,3,2],-1)
([4,1,3,2],1)
([4,3,1,2],-1)
([4,3,2,1],1)```

## Icon and Unicon

Works in both languages.

Translation of: Python
`procedure main(A)    every write("Permutations of length ",n := !A) do       every p := permute(n) do write("\t",showList(p[1])," -> ",right(p[2],2))end procedure permute(n)    items := [[]]    every (j := 1 to n, new_items := []) do {        every item := items[i := 1 to *items] do {            if *item = 0 then put(new_items, [j])            else if i%2 = 0 then                every k := 1 to *item+1 do {                    new_item := item[1:k] ||| [j] ||| item[k:0]                    put(new_items, new_item)                    }            else                every k := *item+1 to 1 by -1 do {                    new_item := item[1:k] ||| [j] ||| item[k:0]                    put(new_items, new_item)                    }            }       items := new_items       }    suspend (i := 0, [!items, if (i+:=1)%2 = 0 then 1 else -1])end procedure showList(A)    every (s := "[") ||:= image(!A)||", "    return s[1:-2]||"]"end`

Sample run:

```->pbs 3 4
Permutations of length 3
[1, 2, 3] -> -1
[1, 3, 2] ->  1
[3, 1, 2] -> -1
[3, 2, 1] ->  1
[2, 3, 1] -> -1
[2, 1, 3] ->  1
Permutations of length 4
[1, 2, 3, 4] -> -1
[1, 2, 4, 3] ->  1
[1, 4, 2, 3] -> -1
[4, 1, 2, 3] ->  1
[4, 1, 3, 2] -> -1
[1, 4, 3, 2] ->  1
[1, 3, 4, 2] -> -1
[1, 3, 2, 4] ->  1
[3, 1, 2, 4] -> -1
[3, 1, 4, 2] ->  1
[3, 4, 1, 2] -> -1
[4, 3, 1, 2] ->  1
[4, 3, 2, 1] -> -1
[3, 4, 2, 1] ->  1
[3, 2, 4, 1] -> -1
[3, 2, 1, 4] ->  1
[2, 3, 1, 4] -> -1
[2, 3, 4, 1] ->  1
[2, 4, 3, 1] -> -1
[4, 2, 3, 1] ->  1
[4, 2, 1, 3] -> -1
[2, 4, 1, 3] ->  1
[2, 1, 4, 3] -> -1
[2, 1, 3, 4] ->  1
->
```

## J

J has a built in mechanism for representing permutations for selecting a permutation of a given length with an integer, but this mechanism does not seem to have an obvious mapping to Steinhaus–Johnson–Trotter. Perhaps someone with a sufficiently deep view of the subject of permutations can find a direct mapping?

Meanwhile, here's an inductive approach, using negative integers to look left and positive integers to look right:

`bfsjt0=: _1 - i.lookingat=: 0 >. <:@# <. [email protected]# + * next=: | >./@:* | > | {~ lookingatbfsjtn=: (((] <@, ] + *@{~) | i. next) C. ] * _1 ^ next < |)^:(*@next)`

Here, bfsjt0 N gives the initial permutation of order N, and bfsjtn^:M bfsjt0 N gives the Mth Steinhaus–Johnson–Trotter permutation of order N. (bf stands for "brute force".)

To convert from the Steinhaus–Johnson–Trotter representation of a permutation to J's representation, use <:@|, or to find J's anagram index of a Steinhaus–Johnson–Trotter representation of a permutation, use [email protected]:<:@:|

Example use:

`   bfsjtn^:(i.!3) bfjt0 3_1 _2 _3_1 _3 _2_3 _1 _2 3 _2 _1_2  3 _1_2 _1  3   <:@| bfsjtn^:(i.!3) bfjt0 30 1 20 2 12 0 12 1 01 2 01 0 2   A. <:@| bfsjtn^:(i.!3) bfjt0 30 1 4 5 3 2`

Here's an example of the Steinhaus–Johnson–Trotter representation of 3 element permutation, with sign (sign is the first column):

`   (_1^2|i.!3),. bfsjtn^:(i.!3) bfjt0 3 1 _1 _2 _3_1 _1 _3 _2 1 _3 _1 _2_1  3 _2 _1 1 _2  3 _1_1 _2 _1  3`

Alternatively, J defines C.!.2 as the parity of a permutation:

`   (,.~C.!.2)<:| bfsjtn^:(i.!3) bfjt0 3 1 0 1 2_1 0 2 1 1 2 0 1_1 2 1 0 1 1 2 0_1 1 0 2`

### Recursive Implementation

This is based on the python recursive implementation:

`rsjt=: 3 :0  if. 2>y do. i.2#y  else.  ((!y)\$(,~|.)-.=i.y)#inv!.(y-1)"1 y#rsjt y-1  end.)`

Example use (here, prefixing each row with its parity):

`   (,.~ C.!.2) rsjt 3 1 0 1 2_1 0 2 1 1 2 0 1_1 2 1 0 1 1 2 0_1 1 0 2`

## Java

Heap's Algorithm, recursive and looping implementations

`package org.rosettacode.java; import java.util.Arrays;import java.util.stream.IntStream; public class HeapsAlgorithm { 	public static void main(String[] args) {		Object[] array = IntStream.range(0, 4)				.boxed()				.toArray();		HeapsAlgorithm algorithm = new HeapsAlgorithm();		algorithm.recursive(array);		System.out.println();		algorithm.loop(array);	} 	void recursive(Object[] array) {		recursive(array, array.length, true);	} 	void recursive(Object[] array, int n, boolean plus) {		if (n == 1) {			output(array, plus);		} else {			for (int i = 0; i < n; i++) {				recursive(array, n - 1, i == 0);				swap(array, n % 2 == 0 ? i : 0, n - 1);			}		}	} 	void output(Object[] array, boolean plus) {		System.out.println(Arrays.toString(array) + (plus ? " +1" : " -1"));	} 	void swap(Object[] array, int a, int b) {		Object o = array[a];		array[a] = array[b];		array[b] = o;	} 	void loop(Object[] array) {		loop(array, array.length);	} 	void loop(Object[] array, int n) {		int[] c = new int[n];		output(array, true);		boolean plus = false;		for (int i = 0; i < n; ) {			if (c[i] < i) {				if (i % 2 == 0) {					swap(array, 0, i);				} else {					swap(array, c[i], i);				}				output(array, plus);				plus = !plus;				c[i]++;				i = 0;			} else {				c[i] = 0;				i++;			}		}	}}`
Output:
```[0, 1, 2, 3] +1
[1, 0, 2, 3] -1
[2, 0, 1, 3] +1
[0, 2, 1, 3] -1
[1, 2, 0, 3] +1
[2, 1, 0, 3] -1
[3, 1, 2, 0] +1
[1, 3, 2, 0] -1
[2, 3, 1, 0] +1
[3, 2, 1, 0] -1
[1, 2, 3, 0] +1
[2, 1, 3, 0] -1
[3, 0, 2, 1] +1
[0, 3, 2, 1] -1
[2, 3, 0, 1] +1
[3, 2, 0, 1] -1
[0, 2, 3, 1] +1
[2, 0, 3, 1] -1
[3, 0, 1, 2] +1
[0, 3, 1, 2] -1
[1, 3, 0, 2] +1
[3, 1, 0, 2] -1
[0, 1, 3, 2] +1
[1, 0, 3, 2] -1

[3, 0, 1, 2] +1
[0, 3, 1, 2] -1
[1, 3, 0, 2] +1
[3, 1, 0, 2] -1
[0, 1, 3, 2] +1
[1, 0, 3, 2] -1
[2, 0, 3, 1] +1
[0, 2, 3, 1] -1
[3, 2, 0, 1] +1
[2, 3, 0, 1] -1
[0, 3, 2, 1] +1
[3, 0, 2, 1] -1
[3, 1, 2, 0] +1
[1, 3, 2, 0] -1
[2, 3, 1, 0] +1
[3, 2, 1, 0] -1
[1, 2, 3, 0] +1
[2, 1, 3, 0] -1
[2, 1, 0, 3] +1
[1, 2, 0, 3] -1
[0, 2, 1, 3] +1
[2, 0, 1, 3] -1
[1, 0, 2, 3] +1
[0, 1, 2, 3] -1
```

## jq

Works with: jq version 1.4

Based on the ruby version - the sequence is generated by swapping adjacent elements.

"permutations" generates a stream of arrays of the form [par, perm], where "par" is the parity of the permutation "perm" of the input array. This array may contain any JSON entities, which are regarded as distinct.

`# The helper function, _recurse, is tail-recursive and therefore in# versions of jq with TCO (tail call optimization) there is no# overhead associated with the recursion. def permutations:  def abs: if . < 0 then -. else . end;  def sign: if . < 0 then -1 elif . == 0 then 0 else 1 end;  def swap(i;j): .[i] as \$i | .[i] = .[j] | .[j] = \$i;   # input: [ parity, extendedPermutation]  def _recurse:    .[0] as \$s | .[1] as \$p | ((\$p | length) -1) as \$n    | [ \$s, (\$p[1:] | map(abs)) ],      (reduce range(2; \$n+1) as \$i         (0;          if \$p[\$i] < 0 and -(\$p[\$i]) > (\$p[\$i-1]|abs) and -(\$p[\$i]) > (\$p[.]|abs)          then \$i           else .          end)) as \$k      | (reduce range(1; \$n) as \$i           (\$k;            if \$p[\$i] > 0 and \$p[\$i] > (\$p[\$i+1]|abs) and \$p[\$i] > (\$p[.]|abs)            then \$i             else .            end)) as \$k      | if \$k == 0 then empty        else (reduce range(1; \$n) as \$i	       (\$p;                if (.[\$i]|abs) > (.[\$k]|abs) then .[\$i] *= -1                 else .                end )) as \$p        | (\$k + (\$p[\$k]|sign)) as \$i        | (\$p | swap(\$i; \$k)) as \$p        | [ -(\$s), \$p ] | _recurse        end ;   . as \$in  | length as \$n  | (reduce range(0; \$n+1) as \$i ([]; . + [ -\$i ])) as \$p  # recurse state: [\$s, \$p]  | [ 1, \$p] | _recurse  | .[1] as \$p  | .[1] = reduce range(0; \$n) as \$i ([]; . + [\$in[\$p[\$i]  - 1]]) ; def count(stream): reduce stream as \$x (0; .+1);`

Examples:

`(["a", "b", "c"] | permutations),"There are \(count( [range(1;6)] | permutations )) permutations of 5 items."`
Output:
`\$ jq -c -n -f Permutations_by_swapping.jq[1,["a","b","c"]][-1,["a","c","b"]][1,["c","a","b"]][-1,["c","b","a"]][1,["b","c","a"]][-1,["b","a","c"]] "There are 32 permutations of 5 items."`

## Julia

Nonrecursive (interative):

` function johnsontrottermove!(ints, isleft)    len = length(ints)    function ismobile(pos)        if isleft[pos] && (pos > 1) && (ints[pos-1] < ints[pos])            return true        elseif !isleft[pos] && (pos < len) && (ints[pos+1] < ints[pos])            return true        end        false    end    function maxmobile()        arr = [ints[pos] for pos in 1:len if ismobile(pos)]        if isempty(arr)            0, 0        else            maxmob = maximum(arr)            maxmob, findfirst(x -> x == maxmob, ints)        end    end    function directedswap(pos)        tmp = ints[pos]        tmpisleft = isleft[pos]        if isleft[pos]            ints[pos] = ints[pos-1]; ints[pos-1] = tmp            isleft[pos] = isleft[pos-1]; isleft[pos-1] = tmpisleft        else            ints[pos] = ints[pos+1]; ints[pos+1] = tmp            isleft[pos] = isleft[pos+1]; isleft[pos+1] = tmpisleft        end    end    (moveint, movepos) = maxmobile()    if movepos > 0        directedswap(movepos)        for (i, val) in enumerate(ints)            if val > moveint                isleft[i] = !isleft[i]            end        end        ints, isleft, true    else        ints, isleft, false    endendfunction johnsontrotter(low, high)    ints = collect(low:high)    isleft = [true for i in ints]    firstconfig = copy(ints)    iters = 0    while true        iters += 1        println("\$ints \$(iters & 1 == 1 ? "+1" : "-1")")        if johnsontrottermove!(ints, isleft)[3] == false            break        end    end    println("There were \$iters iterations.")endjohnsontrotter(1,4) `

Recursive (note this uses memory of roughtly (n+1)! bytes, where n is the number of elements, in order to store the accumulated permutations in a list, and so the above, iterative solution is to be preferred for numbers of elements over 9 or so):

` function johnsontrotter(low, high)    function permutelevel(vec)        if length(vec) < 2            return [vec]        end        sequences = []        endint = vec[end]        smallersequences = permutelevel(vec[1:end-1])        leftward = true        for seq in smallersequences            for pos in (leftward ? (length(seq)+1:-1:1): (1:length(seq)+1))                push!(sequences, insert!(copy(seq), pos, endint))            end            leftward = !leftward        end        sequences    end    permutelevel(collect(low:high))end for (i, sequence) in enumerate(johnsontrotter(1,4))    println("""\$sequence, \$(i & 1 == 1 ? "+1" : "-1")""")end `

## Kotlin

This is based on the recursive Java code found at http://introcs.cs.princeton.edu/java/23recursion/JohnsonTrotter.java.html

`// version 1.1.2 fun johnsonTrotter(n: Int): Pair<List<IntArray>, List<Int>> {    val p = IntArray(n) { it }  // permutation    val q = IntArray(n) { it }  // inverse permutation    val d = IntArray(n) { -1 }  // direction = 1 or -1    var sign = 1    val perms = mutableListOf<IntArray>()    val signs = mutableListOf<Int>()     fun permute(k: Int) {        if (k >= n) {            perms.add(p.copyOf())            signs.add(sign)            sign *= -1            return        }         permute(k + 1)        for (i in 0 until k) {            val z = p[q[k] + d[k]]            p[q[k]] = z            p[q[k] + d[k]] = k            q[z] = q[k]            q[k] += d[k]            permute(k + 1)        }        d[k] *= -1    }      permute(0)    return perms to signs} fun printPermsAndSigns(perms: List<IntArray>, signs: List<Int>) {    for ((i, perm) in perms.withIndex()) {        println("\${perm.contentToString()} -> sign = \${signs[i]}")    }} fun main(args: Array<String>) {    val (perms, signs) = johnsonTrotter(3)    printPermsAndSigns(perms, signs)    println()    val (perms2, signs2) = johnsonTrotter(4)    printPermsAndSigns(perms2, signs2)}`
Output:
```[0, 1, 2] -> sign = 1
[0, 2, 1] -> sign = -1
[2, 0, 1] -> sign = 1
[2, 1, 0] -> sign = -1
[1, 2, 0] -> sign = 1
[1, 0, 2] -> sign = -1

[0, 1, 2, 3] -> sign = 1
[0, 1, 3, 2] -> sign = -1
[0, 3, 1, 2] -> sign = 1
[3, 0, 1, 2] -> sign = -1
[3, 0, 2, 1] -> sign = 1
[0, 3, 2, 1] -> sign = -1
[0, 2, 3, 1] -> sign = 1
[0, 2, 1, 3] -> sign = -1
[2, 0, 1, 3] -> sign = 1
[2, 0, 3, 1] -> sign = -1
[2, 3, 0, 1] -> sign = 1
[3, 2, 0, 1] -> sign = -1
[3, 2, 1, 0] -> sign = 1
[2, 3, 1, 0] -> sign = -1
[2, 1, 3, 0] -> sign = 1
[2, 1, 0, 3] -> sign = -1
[1, 2, 0, 3] -> sign = 1
[1, 2, 3, 0] -> sign = -1
[1, 3, 2, 0] -> sign = 1
[3, 1, 2, 0] -> sign = -1
[3, 1, 0, 2] -> sign = 1
[1, 3, 0, 2] -> sign = -1
[1, 0, 3, 2] -> sign = 1
[1, 0, 2, 3] -> sign = -1
```

## Lua

Translation of: C++
`_JT={}function JT(dim)  local n={ values={}, positions={}, directions={}, sign=1 }  setmetatable(n,{__index=_JT})  for i=1,dim do    n.values[i]=i    n.positions[i]=i    n.directions[i]=-1  end  return nend function _JT:largestMobile()  for i=#self.values,1,-1 do    local loc=self.positions[i]+self.directions[i]    if loc >= 1 and loc <= #self.values and self.values[loc] < i then      return i    end  end  return 0end function _JT:next()  local r=self:largestMobile()  if r==0 then return false end  local rloc=self.positions[r]  local lloc=rloc+self.directions[r]  local l=self.values[lloc]  self.values[lloc],self.values[rloc] = self.values[rloc],self.values[lloc]  self.positions[l],self.positions[r] = self.positions[r],self.positions[l]  self.sign=-self.sign  for i=r+1,#self.directions do self.directions[i]=-self.directions[i] end  return trueend   -- test perm=JT(4)repeat  print(unpack(perm.values))until not perm:next()`
Output:
```1       2       3       4
1       2       4       3
1       4       2       3
4       1       2       3
4       1       3       2
1       4       3       2
1       3       4       2
1       3       2       4
3       1       2       4
3       1       4       2
3       4       1       2
4       3       1       2
4       3       2       1
3       4       2       1
3       2       4       1
3       2       1       4
2       3       1       4
2       3       4       1
2       4       3       1
4       2       3       1
4       2       1       3
2       4       1       3
2       1       4       3
2       1       3       4```

### Coroutine Implementation

This is adapted from the Lua Book .

`local wrap, yield = coroutine.wrap, coroutine.yieldlocal function perm(n)    local r = {}    for i=1,n do r[i]=i end        local sign = 1  return wrap(function()    local function swap(m)            if m==0 then          sign = -sign, yield(sign,r)       else        for i=m,1,-1 do          r[i],r[m]=r[m],r[i]          swap(m-1)          r[i],r[m]=r[m],r[i]        end          end    end    swap(n)  end)endfor sign,r in perm(3) do print(sign,table.unpack(r))end`

## Mathematica

### Recursive

`perms[0] = {{{}, 1}}; perms[n_] :=  Flatten[If[#2 == 1, Reverse, # &]@     Table[{Insert[#1, n, i], (-1)^(n + i) #2}, {i, n}] & @@@    perms[n - 1], 1];`

Example:

`Print["Perm: ", #[[1]], " Sign: ", #[[2]]] & /@ [email protected];`
Output:
```Perm: {1,2,3,4} Sign: 1
Perm: {1,2,4,3} Sign: -1
Perm: {1,4,2,3} Sign: 1
Perm: {4,1,2,3} Sign: -1
Perm: {4,1,3,2} Sign: 1
Perm: {1,4,3,2} Sign: -1
Perm: {1,3,4,2} Sign: 1
Perm: {1,3,2,4} Sign: -1
Perm: {3,1,2,4} Sign: 1
Perm: {3,1,4,2} Sign: -1
Perm: {3,4,1,2} Sign: 1
Perm: {4,3,1,2} Sign: -1
Perm: {4,3,2,1} Sign: 1
Perm: {3,4,2,1} Sign: -1
Perm: {3,2,4,1} Sign: 1
Perm: {3,2,1,4} Sign: -1
Perm: {2,3,1,4} Sign: 1
Perm: {2,3,4,1} Sign: -1
Perm: {2,4,3,1} Sign: 1
Perm: {4,2,3,1} Sign: -1
Perm: {4,2,1,3} Sign: 1
Perm: {2,4,1,3} Sign: -1
Perm: {2,1,4,3} Sign: 1
Perm: {2,1,3,4} Sign: -1```

## Nim

`# iterative Boothroyd methoditerator permutations*[T](ys: openarray[T]): tuple[perm: seq[T], sign: int] =  var    d = 1    c = newSeq[int](ys.len)    xs = newSeq[T](ys.len)    sign = 1   for i, y in ys: xs[i] = y  yield (xs, sign)   block outter:    while true:      while d > 1:        dec d        c[d] = 0      while c[d] >= d:        inc d        if d >= ys.len: break outter       let i = if (d and 1) == 1: c[d] else: 0      swap xs[i], xs[d]      sign *= -1      yield (xs, sign)      inc c[d] if isMainModule:  for i in permutations([0,1,2]):    echo i   echo ""   for i in permutations([0,1,2,3]):    echo i`
Output:
```(perm: @[0, 1, 2], sign: 1)
(perm: @[1, 0, 2], sign: -1)
(perm: @[2, 0, 1], sign: 1)
(perm: @[0, 2, 1], sign: -1)
(perm: @[1, 2, 0], sign: 1)
(perm: @[2, 1, 0], sign: -1)

(perm: @[0, 1, 2, 3], sign: 1)
(perm: @[1, 0, 2, 3], sign: -1)
(perm: @[2, 0, 1, 3], sign: 1)
(perm: @[0, 2, 1, 3], sign: -1)
(perm: @[1, 2, 0, 3], sign: 1)
(perm: @[2, 1, 0, 3], sign: -1)
(perm: @[3, 1, 0, 2], sign: 1)
(perm: @[1, 3, 0, 2], sign: -1)
(perm: @[0, 3, 1, 2], sign: 1)
(perm: @[3, 0, 1, 2], sign: -1)
(perm: @[1, 0, 3, 2], sign: 1)
(perm: @[0, 1, 3, 2], sign: -1)
(perm: @[0, 2, 3, 1], sign: 1)
(perm: @[2, 0, 3, 1], sign: -1)
(perm: @[3, 0, 2, 1], sign: 1)
(perm: @[0, 3, 2, 1], sign: -1)
(perm: @[2, 3, 0, 1], sign: 1)
(perm: @[3, 2, 0, 1], sign: -1)
(perm: @[3, 2, 1, 0], sign: 1)
(perm: @[2, 3, 1, 0], sign: -1)
(perm: @[1, 3, 2, 0], sign: 1)
(perm: @[3, 1, 2, 0], sign: -1)
(perm: @[2, 1, 3, 0], sign: 1)
(perm: @[1, 2, 3, 0], sign: -1)```

## Perl

### S-J-T Based

` #!perluse strict;use warnings; # This code uses "Even's Speedup," as described on# the Wikipedia page about the Steinhaus–Johnson–# Trotter algorithm. # Any resemblance between this code and the Python# code elsewhere on the page is purely a coincidence,# caused by them both implementing the same algorithm. # The code was written to be read relatively easily# while demonstrating some common perl idioms. sub perms(&@) {   my \$callback = shift;   my @perm = map [\$_, -1], @_;   \$perm[0][1] = 0;    my \$sign = 1;   while( ) {      \$callback->(\$sign, map \$_->[0], @perm);      \$sign *= -1;       my (\$chosen, \$index) = (-1, -1);      for my \$i ( 0 .. \$#perm ) {         (\$chosen, \$index) = (\$perm[\$i][0], \$i)           if \$perm[\$i][1] and \$perm[\$i][0] > \$chosen;      }      return if \$index == -1;       my \$direction = \$perm[\$index][1];      my \$next = \$index + \$direction;       @perm[ \$index, \$next ] = @perm[ \$next, \$index ];       if( \$next <= 0 or \$next >= \$#perm ) {         \$perm[\$next][1] = 0;      } elsif( \$perm[\$next + \$direction][0] > \$chosen ) {         \$perm[\$next][1] = 0;      }       for my \$i ( 0 .. \$next - 1 ) {         \$perm[\$i][1] = +1 if \$perm[\$i][0] > \$chosen;      }      for my \$i ( \$next + 1 .. \$#perm ) {         \$perm[\$i][1] = -1 if \$perm[\$i][0] > \$chosen;      }   }} my \$n = shift(@ARGV) || 4; perms {   my (\$sign, @perm) = @_;   print "[", join(", ", @perm), "]";   print \$sign < 0 ? " => -1\n" : " => +1\n";   } 1 .. \$n; `
Output:
```
[1, 2, 3, 4] => +1
[1, 2, 4, 3] => -1
[1, 4, 2, 3] => +1
[4, 1, 2, 3] => -1
[4, 1, 3, 2] => +1
[1, 4, 3, 2] => -1
[1, 3, 4, 2] => +1
[1, 3, 2, 4] => -1
[3, 1, 2, 4] => +1
[3, 1, 4, 2] => -1
[3, 4, 1, 2] => +1
[4, 3, 1, 2] => -1
[4, 3, 2, 1] => +1
[3, 4, 2, 1] => -1
[3, 2, 4, 1] => +1
[3, 2, 1, 4] => -1
[2, 3, 1, 4] => +1
[2, 3, 4, 1] => -1
[2, 4, 3, 1] => +1
[4, 2, 3, 1] => -1
[4, 2, 1, 3] => +1
[2, 4, 1, 3] => -1
[2, 1, 4, 3] => +1
[2, 1, 3, 4] => -1

```

### Alternative Iterative version

This is based on the perl6 recursive version, but without recursion.

`#!perluse strict;use warnings; sub perms {   my (\$xx) = (shift);   my @perms = ([+1]);   for my \$x ( 1 .. \$xx ) {      my \$sign = -1;      @perms = map {         my (\$s, @p) = @\$_;         map [\$sign *= -1, @p[0..\$_-1], \$x, @p[\$_..\$#p]],            \$s < 0 ? 0 .. @p : reverse 0 .. @p;      } @perms;   }   @perms;} my \$n = shift() || 4; for( perms(\$n) ) {   my \$s = shift @\$_;   \$s = '+1' if \$s > 0;   print "[", join(", ", @\$_), "] => \$s\n";} `
Output:

The output is the same as the first perl solution.

## Perl 6

### Recursive

Works with: rakudo version 2015-09-25
`sub insert(\$x, @xs) { ([flat @xs[0 ..^ \$_], \$x, @xs[\$_ .. *]] for 0 .. +@xs) }sub order(\$sg, @xs) { \$sg > 0 ?? @xs !! @xs.reverse } multi perms([]) {    [] => +1} multi perms([\$x, *@xs]) {    perms(@xs).map({ |order(\$_.value, insert(\$x, \$_.key)) }) Z=> |(+1,-1) xx *} .say for perms([0..2]);`
Output:
```[0 1 2] => 1
[1 0 2] => -1
[1 2 0] => 1
[2 1 0] => -1
[2 0 1] => 1
[0 2 1] => -1```

## Phix

Ad-hoc recursive solution, not (knowingly) based on any given algorithm, but instead on achieving the desired pattern.
Only once finished did I properly grasp that odd/even permutation idea, and that it is very nearly the same algorithm.
Only difference is my version directly calculates where to insert p, without using the parity (which I added in last).

`function spermutations(integer p, integer i)-- generate the i'th permutation of [1..p]:-- first obtain the appropriate permutation of [1..p-1],-- then insert p/move it down k(=0..p-1) places from the end.    integer k = mod(i-1,2*p)    if k>=p then k=2*p-1-k  end if    sequence res    integer parity    if p>1 then        {res,parity} = spermutations(p-1,floor((i-1)/p)+1)        res = res[1..length(res)-k]&p&res[length(res)-k+1..\$]    else        res = {1}    end if    return {res,iff(and_bits(i,1)?1:-1)}end function for p=1 to 4 do    printf(1,"==%d==\n",p)    for i=1 to factorial(p) do        ?{i,spermutations(p,i)}    end forend for`
Output:
```"started"
==1==
{1,{{1},1}}
==2==
{1,{{1,2},1}}
{2,{{2,1},-1}}
==3==
{1,{{1,2,3},1}}
{2,{{1,3,2},-1}}
{3,{{3,1,2},1}}
{4,{{3,2,1},-1}}
{5,{{2,3,1},1}}
{6,{{2,1,3},-1}}
==4==
{1,{{1,2,3,4},1}}
{2,{{1,2,4,3},-1}}
{3,{{1,4,2,3},1}}
{4,{{4,1,2,3},-1}}
{5,{{4,1,3,2},1}}
{6,{{1,4,3,2},-1}}
{7,{{1,3,4,2},1}}
{8,{{1,3,2,4},-1}}
{9,{{3,1,2,4},1}}
{10,{{3,1,4,2},-1}}
{11,{{3,4,1,2},1}}
{12,{{4,3,1,2},-1}}
{13,{{4,3,2,1},1}}
{14,{{3,4,2,1},-1}}
{15,{{3,2,4,1},1}}
{16,{{3,2,1,4},-1}}
{17,{{2,3,1,4},1}}
{18,{{2,3,4,1},-1}}
{19,{{2,4,3,1},1}}
{20,{{4,2,3,1},-1}}
{21,{{4,2,1,3},1}}
{22,{{2,4,1,3},-1}}
{23,{{2,1,4,3},1}}
{24,{{2,1,3,4},-1}}
```

## PicoLisp

`(let   (N 4      L      (mapcar         '((I) (list I 0))         (range 1 N) ) )   (for I L      (printsp (car I)) )   (prinl)   (while      # find the lagest mobile integer      (setq         X         (maxi            '((I) (car (get L (car I))))            (extract               '((I J)                  (let? Y                     (get                        L                        ((if (=0 (cadr I)) dec inc) J) )                     (when (> (car I) (car Y))                        (list J (cadr I)) ) ) )               L               (range 1 N) ) )         Y (get L (car X)) )      # swap integer and adjacent int it is looking at      (xchg         (nth L (car X))         (nth            L            ((if (=0 (cadr X)) dec inc) (car X)) ) )      # reverse direction of all ints large than our      (for I L         (when (< (car Y) (car I))            (set (cdr I)               (if (=0 (cadr I)) 1 0) ) ) )      # print current positions      (for I L         (printsp (car I)) )      (prinl) ) )(bye)`

## PowerShell

` function permutation (\$array) {    function sign(\$A) {        \$size = \$A.Count        \$sign = 1        for(\$i = 0; \$i -lt \$size; \$i++) {            for(\$j = \$i+1; \$j -lt \$size ; \$j++) {                if(\$A[\$j] -lt \$A[\$i]) { \$sign *= -1}            }        }        \$sign    }    function generate(\$n, \$A, \$i1, \$i2, \$cnt) {        if(\$n -eq 1) {            if(\$cnt -gt 0) {                "\$A -- swapped positions: \$i1 \$i2 -- sign = \$(sign \$A)`n"            } else {                "\$A -- sign = \$(sign \$A)`n"            }        }        else{            for( \$i = 0; \$i -lt (\$n - 1); \$i += 1) {                generate (\$n - 1) \$A \$i1 \$i2 \$cnt                if(\$n % 2 -eq 0){                    \$i1, \$i2 = \$i, (\$n-1)                    \$A[\$i1], \$A[\$i2] = \$A[\$i2], \$A[\$i1]                    \$cnt = 1                }                else{                    \$i1, \$i2 = 0, (\$n-1)                    \$A[\$i1], \$A[\$i2] = \$A[\$i2], \$A[\$i1]                    \$cnt = 1                }            }            generate (\$n - 1) \$A \$i1 \$i2 \$cnt        }    }    \$n = \$array.Count    if(\$n -gt 0) {        (generate \$n \$array  0 (\$n-1) 0)    } else {\$array}}permutation @(1,2,3,4) `

Output:

```1 2 3 4 -- sign = 1

2 1 3 4 -- swapped positions: 0 1 -- sign = -1

3 1 2 4 -- swapped positions: 0 2 -- sign = 1

1 3 2 4 -- swapped positions: 0 1 -- sign = -1

2 3 1 4 -- swapped positions: 0 2 -- sign = 1

3 2 1 4 -- swapped positions: 0 1 -- sign = -1

4 2 1 3 -- swapped positions: 0 3 -- sign = 1

2 4 1 3 -- swapped positions: 0 1 -- sign = -1

1 4 2 3 -- swapped positions: 0 2 -- sign = 1

4 1 2 3 -- swapped positions: 0 1 -- sign = -1

2 1 4 3 -- swapped positions: 0 2 -- sign = 1

1 2 4 3 -- swapped positions: 0 1 -- sign = -1

1 3 4 2 -- swapped positions: 1 3 -- sign = 1

3 1 4 2 -- swapped positions: 0 1 -- sign = -1

4 1 3 2 -- swapped positions: 0 2 -- sign = 1

1 4 3 2 -- swapped positions: 0 1 -- sign = -1

3 4 1 2 -- swapped positions: 0 2 -- sign = 1

4 3 1 2 -- swapped positions: 0 1 -- sign = -1

4 3 2 1 -- swapped positions: 2 3 -- sign = 1

3 4 2 1 -- swapped positions: 0 1 -- sign = -1

2 4 3 1 -- swapped positions: 0 2 -- sign = 1

4 2 3 1 -- swapped positions: 0 1 -- sign = -1

3 2 4 1 -- swapped positions: 0 2 -- sign = 1

2 3 4 1 -- swapped positions: 0 1 -- sign = -1
```

## Python

### Python: iterative

When saved in a file called spermutations.py it is used in the Python example to the Matrix arithmetic task and so any changes here should also be reflected and checked in that task example too.

`from operator import itemgetter DEBUG = False # like the built-in __debug__ def spermutations(n):    """permutations by swapping. Yields: perm, sign"""    sign = 1    p = [[i, 0 if i == 0 else -1] # [num, direction]         for i in range(n)]     if DEBUG: print ' #', p    yield tuple(pp[0] for pp in p), sign     while any(pp[1] for pp in p): # moving        i1, (n1, d1) = max(((i, pp) for i, pp in enumerate(p) if pp[1]),                           key=itemgetter(1))        sign *= -1        if d1 == -1:            # Swap down            i2 = i1 - 1            p[i1], p[i2] = p[i2], p[i1]            # If this causes the chosen element to reach the First or last            # position within the permutation, or if the next element in the            # same direction is larger than the chosen element:            if i2 == 0 or p[i2 - 1][0] > n1:                # The direction of the chosen element is set to zero                p[i2][1] = 0        elif d1 == 1:            # Swap up            i2 = i1 + 1            p[i1], p[i2] = p[i2], p[i1]            # If this causes the chosen element to reach the first or Last            # position within the permutation, or if the next element in the            # same direction is larger than the chosen element:            if i2 == n - 1 or p[i2 + 1][0] > n1:                # The direction of the chosen element is set to zero                p[i2][1] = 0        if DEBUG: print ' #', p        yield tuple(pp[0] for pp in p), sign         for i3, pp in enumerate(p):            n3, d3 = pp            if n3 > n1:                pp[1] = 1 if i3 < i2 else -1                if DEBUG: print ' # Set Moving'  if __name__ == '__main__':    from itertools import permutations     for n in (3, 4):        print '\nPermutations and sign of %i items' % n        sp = set()        for i in spermutations(n):            sp.add(i[0])            print('Perm: %r Sign: %2i' % i)            #if DEBUG: raw_input('?')        # Test        p = set(permutations(range(n)))        assert sp == p, 'Two methods of generating permutations do not agree'`
Output:
```Permutations and sign of 3 items
Perm: (0, 1, 2) Sign:  1
Perm: (0, 2, 1) Sign: -1
Perm: (2, 0, 1) Sign:  1
Perm: (2, 1, 0) Sign: -1
Perm: (1, 2, 0) Sign:  1
Perm: (1, 0, 2) Sign: -1

Permutations and sign of 4 items
Perm: (0, 1, 2, 3) Sign:  1
Perm: (0, 1, 3, 2) Sign: -1
Perm: (0, 3, 1, 2) Sign:  1
Perm: (3, 0, 1, 2) Sign: -1
Perm: (3, 0, 2, 1) Sign:  1
Perm: (0, 3, 2, 1) Sign: -1
Perm: (0, 2, 3, 1) Sign:  1
Perm: (0, 2, 1, 3) Sign: -1
Perm: (2, 0, 1, 3) Sign:  1
Perm: (2, 0, 3, 1) Sign: -1
Perm: (2, 3, 0, 1) Sign:  1
Perm: (3, 2, 0, 1) Sign: -1
Perm: (3, 2, 1, 0) Sign:  1
Perm: (2, 3, 1, 0) Sign: -1
Perm: (2, 1, 3, 0) Sign:  1
Perm: (2, 1, 0, 3) Sign: -1
Perm: (1, 2, 0, 3) Sign:  1
Perm: (1, 2, 3, 0) Sign: -1
Perm: (1, 3, 2, 0) Sign:  1
Perm: (3, 1, 2, 0) Sign: -1
Perm: (3, 1, 0, 2) Sign:  1
Perm: (1, 3, 0, 2) Sign: -1
Perm: (1, 0, 3, 2) Sign:  1
Perm: (1, 0, 2, 3) Sign: -1```

### Python: recursive

After spotting the pattern of highest number being inserted into each perm of lower numbers from right to left, then left to right, I developed this recursive function:

`def s_permutations(seq):    def s_perm(seq):        if not seq:            return [[]]        else:            new_items = []            for i, item in enumerate(s_perm(seq[:-1])):                if i % 2:                    # step up                    new_items += [item[:i] + seq[-1:] + item[i:]                                  for i in range(len(item) + 1)]                else:                    # step down                    new_items += [item[:i] + seq[-1:] + item[i:]                                  for i in range(len(item), -1, -1)]            return new_items     return [(tuple(item), -1 if i % 2 else 1)            for i, item in enumerate(s_perm(seq))]`
Sample output:

The output is the same as before except it is a list of all results rather than yielding each result from a generator function.

### Python: Iterative version of the recursive

Replacing the recursion in the example above produces this iterative version function:

`def s_permutations(seq):    items = [[]]    for j in seq:        new_items = []        for i, item in enumerate(items):            if i % 2:                # step up                new_items += [item[:i] + [j] + item[i:]                              for i in range(len(item) + 1)]            else:                # step down                new_items += [item[:i] + [j] + item[i:]                              for i in range(len(item), -1, -1)]        items = new_items     return [(tuple(item), -1 if i % 2 else 1)            for i, item in enumerate(items)]`
Sample output:

The output is the same as before and is a list of all results rather than yielding each result from a generator function.

## Racket

` #lang racket (define (add-at l i x)  (if (zero? i) (cons x l) (cons (car l) (add-at (cdr l) (sub1 i) x)))) (define (permutations l)  (define (loop l)    (cond [(null? l) '(())]          [else (for*/list ([(p i) (in-indexed (loop (cdr l)))]                            [i ((if (odd? i) identity reverse)                                (range (add1 (length p))))])                  (add-at p i (car l)))]))  (for/list ([p (loop (reverse l))] [i (in-cycle '(1 -1))]) (cons i p))) (define (show-permutations l)  (printf "Permutations of ~s:\n" l)  (for ([p (permutations l)])    (printf "  ~a (~a)\n" (apply ~a (add-between (cdr p) ", ")) (car p)))) (for ([n (in-range 3 5)]) (show-permutations (range n))) `
Output:
```Permutations of (0 1 2):
0, 1, 2 (1)
0, 2, 1 (-1)
2, 0, 1 (1)
2, 1, 0 (-1)
1, 2, 0 (1)
1, 0, 2 (-1)
Permutations of (0 1 2 3):
0, 1, 2, 3 (1)
0, 1, 3, 2 (-1)
0, 3, 1, 2 (1)
3, 0, 1, 2 (-1)
3, 0, 2, 1 (1)
0, 3, 2, 1 (-1)
0, 2, 3, 1 (1)
0, 2, 1, 3 (-1)
2, 0, 1, 3 (1)
2, 0, 3, 1 (-1)
2, 3, 0, 1 (1)
3, 2, 0, 1 (-1)
3, 2, 1, 0 (1)
2, 3, 1, 0 (-1)
2, 1, 3, 0 (1)
2, 1, 0, 3 (-1)
1, 2, 0, 3 (1)
1, 2, 3, 0 (-1)
1, 3, 2, 0 (1)
3, 1, 2, 0 (-1)
3, 1, 0, 2 (1)
1, 3, 0, 2 (-1)
1, 0, 3, 2 (1)
1, 0, 2, 3 (-1)
```

## REXX

`/*REXX program  generates all  permutations  of   N   different objects by  swapping.   */parse arg things bunch .                         /*obtain optional arguments from the CL*/if things=='' | things==","  then things=4       /*Not specified?  Then use the default.*/if bunch =='' | bunch ==","  then bunch =things  /* "      "         "   "   "     "    */call permSets things, bunch                      /*invoke permutations by swapping sub. */exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/!:        procedure;  !=1;        do j=2  to arg(1);    !=!*j;     end;           return !/*──────────────────────────────────────────────────────────────────────────────────────*/permSets: procedure; parse arg x,y               /*take   X  things   Y   at a time.    */          !.=0;      pad=left('', x*y)           /*X can't be > length of below str (62)*/          z=left('123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', x);  q=z          #=1                                    /*the number of permutations  (so far).*/          !.z=1;    s=1;   times=!(x) % !(x-y)   /*calculate (#) TIMES  using factorial.*/          w=max(length(z), length('permute') )   /*maximum width of  Z and also PERMUTE.*/          say center('permutations for '   x   ' things taken '   y   " at a time",60,'═')          say          say   pad    'permutation'       center("permute", w, '─')         "sign"          say   pad    '───────────'       center("───────", w, '─')         "────"          say   pad    center(#, 11)       center(z        , w)              right(s, 4-1)              do \$=1   until  #==times            /*perform permutation until # of times.*/               do   k=1    for x-1               /*step thru things for  things-1 times.*/                 do m=k+1  to  x;      ?=        /*this method doesn't use  adjacency.  */                     do n=1  for x               /*build the new permutation by swapping*/                     if n\==k & n\==m  then               ? =  ?  ||  substr(z, n, 1)                                       else if n==k  then ? =  ?  ||  substr(z, m, 1)                                                     else ? =  ?  ||  substr(z, k, 1)                     end   /*n*/                 z=?                             /*save this permutation for next swap. */                 if !.?  then iterate m          /*if defined before, then try next one.*/                 _=0                             /* [↓]  count number of swapped symbols*/                    do d=1  for x  while \$\==1;  _= _ + (substr(?,d,1)\==substr(prev,d,1))                    end   /*d*/                 if _>2  then do;        _=z                              a=\$//x+1;  q=q + _ /* [← ↓]  this swapping tries adjacency*/                              b=q//x+1;  if b==a  then b=a + 1;       if b>x  then b=a - 1                              z=overlay( substr(z,b,1), overlay( substr(z,a,1), _, b),  a)                              iterate \$          /*now, try this particular permutation.*/                              end                 #=#+1;  s= -s;   say pad   center(#, 11)    center(?, w)    right(s, 4-1)                 !.?=1;  prev=?;      iterate \$  /*now, try another swapped permutation.*/                 end   /*m*/               end     /*k*/             end       /*\$*/          return                                 /*we're all finished with permutating. */`
output   when using the default input:
```══════permutations for  4  things taken  4  at a time═══════

permutation permute sign
─────────── ─────── ────
1       1234     1
2       2134    -1
3       3124     1
4       1324    -1
5       1342     1
6       3142    -1
7       4132     1
8       1432    -1
9       2431     1
10       4231    -1
11       4321     1
12       3421    -1
13       3241     1
14       2341    -1
15       2314     1
16       3214    -1
17       3412     1
18       4312    -1
19       4213     1
20       2413    -1
21       2143     1
22       1243    -1
23       1423     1
24       4123    -1
```

## Ruby

Translation of: BBC BASIC
`def perms(n)  p = Array.new(n+1){|i| -i}  s = 1  loop do    yield p[1..-1].map(&:abs), s    k = 0    for i in 2..n      k = i if p[i] < 0 and p[i].abs > p[i-1].abs and p[i].abs > p[k].abs    end    for i in 1...n      k = i if p[i] > 0 and p[i].abs > p[i+1].abs and  p[i].abs > p[k].abs    end    break if k.zero?    for i in 1..n      p[i] *= -1 if p[i].abs > p[k].abs    end    i = k + (p[k] <=> 0)    p[k], p[i] = p[i], p[k]    s = -s  endend for i in 3..4  perms(i){|perm, sign| puts "Perm: #{perm}  Sign: #{sign}"}  putsend`
Output:
```Perm: [1, 2, 3]  Sign: 1
Perm: [1, 3, 2]  Sign: -1
Perm: [3, 1, 2]  Sign: 1
Perm: [3, 2, 1]  Sign: -1
Perm: [2, 3, 1]  Sign: 1
Perm: [2, 1, 3]  Sign: -1

Perm: [1, 2, 3, 4]  Sign: 1
Perm: [1, 2, 4, 3]  Sign: -1
Perm: [1, 4, 2, 3]  Sign: 1
Perm: [4, 1, 2, 3]  Sign: -1
Perm: [4, 1, 3, 2]  Sign: 1
Perm: [1, 4, 3, 2]  Sign: -1
Perm: [1, 3, 4, 2]  Sign: 1
Perm: [1, 3, 2, 4]  Sign: -1
Perm: [3, 1, 2, 4]  Sign: 1
Perm: [3, 1, 4, 2]  Sign: -1
Perm: [3, 4, 1, 2]  Sign: 1
Perm: [4, 3, 1, 2]  Sign: -1
Perm: [4, 3, 2, 1]  Sign: 1
Perm: [3, 4, 2, 1]  Sign: -1
Perm: [3, 2, 4, 1]  Sign: 1
Perm: [3, 2, 1, 4]  Sign: -1
Perm: [2, 3, 1, 4]  Sign: 1
Perm: [2, 3, 4, 1]  Sign: -1
Perm: [2, 4, 3, 1]  Sign: 1
Perm: [4, 2, 3, 1]  Sign: -1
Perm: [4, 2, 1, 3]  Sign: 1
Perm: [2, 4, 1, 3]  Sign: -1
Perm: [2, 1, 4, 3]  Sign: 1
Perm: [2, 1, 3, 4]  Sign: -1
```

## Scala

`object JohnsonTrotter extends App {   private def perm(n: Int): Unit = {    val p = new Array[Int](n) // permutation    val pi = new Array[Int](n) // inverse permutation    val dir = new Array[Int](n) // direction = +1 or -1     def perm(n: Int, p: Array[Int], pi: Array[Int], dir: Array[Int]): Unit = {      if (n >= p.length) for (aP <- p) print(aP)      else {        perm(n + 1, p, pi, dir)        for (i <- 0 until n) { // swap          printf("   (%d %d)\n", pi(n), pi(n) + dir(n))          val z = p(pi(n) + dir(n))          p(pi(n)) = z          p(pi(n) + dir(n)) = n          pi(z) = pi(n)          pi(n) = pi(n) + dir(n)          perm(n + 1, p, pi, dir)        }        dir(n) = -dir(n)      }    }     for (i <- 0 until n) {      dir(i) = -1      p(i) = i      pi(i) = i    }    perm(0, p, pi, dir)    print("   (0 1)\n")  }   perm(4) }`
Output:
See it in running in your browser by Scastie (JVM).

## Sidef

Translation of: Perl
`func perms(n) {   var perms = [[+1]]   for x in (1..n) {      var sign = -1      perms = gather {        for s,*p in perms {          var r = (0 .. p.len)          take((s < 0 ? r : r.flip).map {|i|            [sign *= -1, p[^i], x, p[i..p.end]]          }...)        }      }   }   perms} var n = 4for p in perms(n) {    var s = p.shift    s > 0 && (s = '+1')    say "#{p} => #{s}"}`
Output:
```[1, 2, 3, 4] => +1
[1, 2, 4, 3] => -1
[1, 4, 2, 3] => +1
[4, 1, 2, 3] => -1
[4, 1, 3, 2] => +1
[1, 4, 3, 2] => -1
[1, 3, 4, 2] => +1
[1, 3, 2, 4] => -1
[3, 1, 2, 4] => +1
[3, 1, 4, 2] => -1
[3, 4, 1, 2] => +1
[4, 3, 1, 2] => -1
[4, 3, 2, 1] => +1
[3, 4, 2, 1] => -1
[3, 2, 4, 1] => +1
[3, 2, 1, 4] => -1
[2, 3, 1, 4] => +1
[2, 3, 4, 1] => -1
[2, 4, 3, 1] => +1
[4, 2, 3, 1] => -1
[4, 2, 1, 3] => +1
[2, 4, 1, 3] => -1
[2, 1, 4, 3] => +1
[2, 1, 3, 4] => -1
```

## Tcl

`# A simple swap operationproc swap {listvar i1 i2} {    upvar 1 \$listvar l    set tmp [lindex \$l \$i1]    lset l \$i1 [lindex \$l \$i2]    lset l \$i2 \$tmp} proc permswap {n v1 v2 body} {    upvar 1 \$v1 perm \$v2 sign     # Initialize    set sign -1    for {set i 0} {\$i < \$n} {incr i} {	lappend items \$i	lappend dirs -1    }     while 1 {	# Report via callback	set perm \$items	set sign [expr {-\$sign}]	uplevel 1 \$body 	# Find the largest mobile integer (lmi) and its index (idx)	set i [set idx -1]	foreach item \$items dir \$dirs {	    set j [expr {[incr i] + \$dir}]	    if {\$j < 0 || \$j >= [llength \$items]} continue	    if {\$item > [lindex \$items \$j] && (\$idx == -1 || \$item > \$lmi)} {		set lmi \$item		set idx \$i	    }	} 	# If none, we're done	if {\$idx == -1} break 	# Swap the largest mobile integer with "what it is looking at"	set nextIdx [expr {\$idx + [lindex \$dirs \$idx]}]	swap items \$idx \$nextIdx	swap dirs \$idx \$nextIdx 	# Reverse directions on larger integers	set i -1	foreach item \$items dir \$dirs {	    lset dirs [incr i] [expr {\$item > \$lmi ? -\$dir : \$dir}]	}    }}`

Demonstrating:

`permswap 4 p s {    puts "\$s\t\$p"}`
Output:
```1	0 1 2 3
-1	0 1 3 2
1	0 3 1 2
-1	3 0 1 2
1	3 0 2 1
-1	0 3 2 1
1	0 2 3 1
-1	0 2 1 3
1	2 0 1 3
-1	2 0 3 1
1	2 3 0 1
-1	3 2 0 1
1	3 2 1 0
-1	2 3 1 0
1	2 1 3 0
-1	2 1 0 3
1	1 2 0 3
-1	1 2 3 0
1	1 3 2 0
-1	3 1 2 0
1	3 1 0 2
-1	1 3 0 2
1	1 0 3 2
-1	1 0 2 3
```

## XPL0

Translation of BBC BASIC example, which uses the Johnson-Trotter algorithm.

`include c:\cxpl\codes; proc PERMS(N);int  N;                         \number of elementsint  I, K, S, T, P;[P:= Reserve((N+1)*4);for I:= 0 to N do P(I):= -I;    \initialize facing left (also set P(0)=0)S:= 1;repeat  Text(0, "Perm: [ ");        for I:= 1 to N do                [IntOut(0, abs(P(I)));  ChOut(0, ^ )];        Text(0, "] Sign: ");  IntOut(0, S);  CrLf(0);         K:= 0;                  \find largest mobile element        for I:= 2 to N do                         \for left-facing elements            if P(I) < 0 and                abs(P(I)) > abs(P(I-1)) and       \ greater than neighbor                abs(P(I)) > abs(P(K)) then K:= I; \ get largest element        for I:= 1 to N-1 do                       \for right-facing elements            if P(I) > 0 and                abs(P(I)) > abs(P(I+1)) and       \ greater than neighbor                abs(P(I)) > abs(P(K)) then K:= I; \ get largest element        if K # 0 then           \mobile element found           [for I:= 1 to N do   \reverse elements > K                if abs(P(I)) > abs(P(K)) then P(I):= P(I)*-1;            I:= K + (if P(K)<0 then -1 else 1);            T:= P(K);  P(K):= P(I);  P(I):= T;    \swap K with element looked at            S:= -S;             \alternate signs            ];until   K = 0;                  \no mobile element remains]; [PERMS(3);CrLf(0);PERMS(4);]`
Output:
```Perm: [ 1 2 3 ] Sign: 1
Perm: [ 1 3 2 ] Sign: -1
Perm: [ 3 1 2 ] Sign: 1
Perm: [ 3 2 1 ] Sign: -1
Perm: [ 2 3 1 ] Sign: 1
Perm: [ 2 1 3 ] Sign: -1

Perm: [ 1 2 3 4 ] Sign: 1
Perm: [ 1 2 4 3 ] Sign: -1
Perm: [ 1 4 2 3 ] Sign: 1
Perm: [ 4 1 2 3 ] Sign: -1
Perm: [ 4 1 3 2 ] Sign: 1
Perm: [ 1 4 3 2 ] Sign: -1
Perm: [ 1 3 4 2 ] Sign: 1
Perm: [ 1 3 2 4 ] Sign: -1
Perm: [ 3 1 2 4 ] Sign: 1
Perm: [ 3 1 4 2 ] Sign: -1
Perm: [ 3 4 1 2 ] Sign: 1
Perm: [ 4 3 1 2 ] Sign: -1
Perm: [ 4 3 2 1 ] Sign: 1
Perm: [ 3 4 2 1 ] Sign: -1
Perm: [ 3 2 4 1 ] Sign: 1
Perm: [ 3 2 1 4 ] Sign: -1
Perm: [ 2 3 1 4 ] Sign: 1
Perm: [ 2 3 4 1 ] Sign: -1
Perm: [ 2 4 3 1 ] Sign: 1
Perm: [ 4 2 3 1 ] Sign: -1
Perm: [ 4 2 1 3 ] Sign: 1
Perm: [ 2 4 1 3 ] Sign: -1
Perm: [ 2 1 4 3 ] Sign: 1
Perm: [ 2 1 3 4 ] Sign: -1
```

## zkl

Translation of: Python
`fcn permute(seq){   insertEverywhere := fcn(x,list){ //(x,(a,b))-->((x,a,b),(a,x,b),(a,b,x))      (0).pump(list.len()+1,List,'wrap(n){list[0,n].extend(x,list[n,*]) })};   insertEverywhereB := fcn(x,t){ //--> insertEverywhere().reverse()      [t.len()..-1,-1].pump(t.len()+1,List,'wrap(n){t[0,n].extend(x,t[n,*])})};    seq.reduce('wrap(items,x){      f := Utils.Helpers.cycle(insertEverywhereB,insertEverywhere);      items.pump(List,'wrap(item){f.next()(x,item)},	      T.fp(Void.Write,Void.Write));   },T(T));}`

A cycle of two "build list" functions is used to insert x forward or reverse. reduce loops over the items and retains the enlarging list of permuations. pump loops over the existing set of permutations and inserts/builds the next set (into a list sink). (Void.Write,Void.Write,list) is a sentinel that says to write the contents of the list to the sink (ie sink.extend(list)). T.fp is a partial application of ROList.create (read only list) and the parameters VW,VW. It will be called (by pump) with a list of lists --> T.create(VM,VM,list) --> list

`p := permute(T(1,2,3));p.println(); p := permute([1..4]);p.len().println();p.toString(*).println()`
Output:
```L(L(1,2,3),L(1,3,2),L(3,1,2),L(3,2,1),L(2,3,1),L(2,1,3))

24
L(
L(1,2,3,4), L(1,2,4,3), L(1,4,2,3), L(4,1,2,3), L(4,1,3,2), L(1,4,3,2),
L(1,3,4,2), L(1,3,2,4), L(3,1,2,4), L(3,1,4,2), L(3,4,1,2), L(4,3,1,2),
L(4,3,2,1), L(3,4,2,1), L(3,2,4,1), L(3,2,1,4), L(2,3,1,4), L(2,3,4,1),
L(2,4,3,1), L(4,2,3,1), L(4,2,1,3), L(2,4,1,3), L(2,1,4,3), L(2,1,3,4) )
```

An iterative, lazy version, which is handy as the number of permutations is n!. Uses "Even's Speedup" as described in the Wikipedia article:

` fcn [private] _permuteW(seq){	// lazy version   N:=seq.len(); NM1:=N-1;   ds:=(0).pump(N,List,T(Void,-1)).copy(); ds[0]=0; // direction to move e: -1,0,1   es:=(0).pump(N,List).copy();  // enumerate seq    while(1) {      vm.yield(es.pump(List,seq.__sGet));       // find biggest e with d!=0      reg i=Void, c=-1;      foreach n in (N){ if(ds[n] and es[n]>c) { c=es[n]; i=n; } }      if(Void==i) return();       d:=ds[i]; j:=i+d;      es.swap(i,j); ds.swap(i,j);	// d tracks e      if(j==NM1 or j==0 or es[j+d]>c) ds[j]=0;      foreach e in (N){ if(es[e]>c) ds[e]=(i-e).sign }   } }  fcn permuteW(seq) { Utils.Generator(_permuteW,seq) }`
`foreach p in (permuteW(T("a","b","c"))){ println(p) }`
Output:
```L("a","b","c")
L("a","c","b")
L("c","a","b")
L("c","b","a")
L("b","c","a")
L("b","a","c")
```